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Ta có: \(\left\{{}\begin{matrix}2\left(x-y\right)+3x=1\\3x+2\left(x-y\right)=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-2y=1\\5x-2y=7\end{matrix}\right.\)(Vô lý)
Vậy: Hệ phương trình vô nghiệm
a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)
$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$
`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$
`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$
Vậy HPT vô nghiệm
1. \(\left\{{}\begin{matrix}3x^2+y^2+4xy=8\left(1\right)\\\left(x+y\right)\left(x^2+xy+2\right)=8\end{matrix}\right.\)
=> \(3x^2+3xy+xy+y^2=\left(x+y\right)\left(x^2+xy+2\right)\)
<=> \(\left(x+y\right)\left(3x+y\right)=\left(x+y\right)\left(x^2+xy+2\right)=0\)
<=> \(\left(x+y\right)\left(x^2+xy+2-3x-y\right)=0\)
<=> \(\left[{}\begin{matrix}x=-y\\x^2+xy+2-3x-y=0\end{matrix}\right.\)
TH1: x = -y thay vào pt (1), ta được:
3y2 + y2 - 4y2 = 8
<=> 0y = 8 (vô lí)
TH2: \(x^2+xy+2-3x-y=0\)
<=> x (x + y) - (x + y) - 2(x - 1) = 0
<=> (x - 1)(x + y) - 2(X - 1) = 0
<=> (x - 1)(x + y - 2) = 0
<=> \(\left[{}\begin{matrix}x=1\\x+y-2=0\end{matrix}\right.\)
Với x = 1 thay vào pt (1) -> 3 + y2 + 4y = 8
<=> y2 + 4y - 5 = 0 <=> (y + 5)(y - 1) = 0
<=> \(\left[{}\begin{matrix}y=-5\\y=1\end{matrix}\right.\)
Với x + y - 2 = 0 => x = 2 - y thay vào pt (1)
=> 3(2 - y)2 + y2 + 4(2 - y)y = 8
<=> 3y2 - 12y + 12 + y2 + 8 - 4y2 = 8
<=> 12 = 12y <=> y= 1 => x = 2 - 1 = 1
Vậy ....
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)
=>xy-2x=xy-4x+2y-8 và 2xy+7x-6y-21=2xy+6x-7y-21
=>2x-2y=-8 và x+y=0
=>x-y=-4 và x+y=0
=>2x=-4 và x+y=0
=>x=-2 và y=2
\(\left\{{}\begin{matrix}y^2=\left(x+7\right)\left(x-2\right)\left(1\right)\\3x^2-4xy+y^2=4\left(1-x\right)\left(2\right)\end{matrix}\right.\)
Từ (2) <=> 3x2 - 4xy + y2 + 4x - 4 = 0
<=> (2x - y)2 - (x2 - 4x + 4) = 0
<=> (2x - y)2 - (x - 2)2 = 0
<=> \(\left(2x-y-x+2\right)\left(2x-y+x-2\right)=0\)
<=> \(\left(x-y+2\right)\left(3x-y-2\right)=0\)
<=> \(\left[{}\begin{matrix}x-y+2=0\\3x-y-2=0\end{matrix}\right.\)
TH1: x - y + 2 = 0 > y = x + 2 thay vào pt (1)
(x + 2)2 = (x + 7)(x - 2)
<=> x2 + 4x + 4 = x2 + 5x - 14
<=> x = 18 => y = 18 + 2 = 20
TH2: 3x - y - 2 = 0 <=> y = 3x - 2 thay vào pt (1)
(3x - 2)2 = (x + 7)(x - 2)
<=> 9x2 - 12x + 4 = x2 + 5x - 14
<=> 8x2 - 17x + 18 = 0
<=> 8(x2 - 17/8x + 289/256) + 287/32 = 0
<=> 8(x - 17/16)2 + 287/32 = 0
=> pt vô nghiệm
Vậy ...